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Moishe Kohan
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First of all, the constants $c_i$ will have to depend on the complex structure of $X$ since without prescribing a complex structure one cannot talks about dependence on a basis of the space of holomorphic differentials on $X$. Now, $h$ and $g$ are both Riemannian metrics on a compact manifold and, thus, the identity map $$ (X,g)\to (X,h) $$ is bi-Lipschitz with Lipschitz constants depending, of course, on $g$ and $h$. Suppose that $(X, h), (Y, h)$$(X, g), (Y, h)$ are Riemannian manifolds and $f: (X, g)\to (Y, h)$ is an $L$-Lipschitz map. Then for every closed geodesic $c$ in $(X,g)$, the $h$-length $\ell_h(c)$ of $f(c)$ is at most $L\ell_g(c)$. Hence, if $c^*$ is a shortest closed geodesic in the free homotopy class of $f(c)$, we have $$ \ell_h(c^*)\le L\ell_g(c). $$ For $L$-bi-Lipschitz maps one obtains the reverse inequality by considering $f^{-1}$. Thus, for every free homotopy class $[c]$ in $X$ you obtain the inequality $$ L^{-1}\ell_g([c]) \le \ell_h([f(c)])\le L \ell_g([c]), $$ where $\ell_\bullet([\bullet])$ stands for the length of a shortest loop in the given free homotopy class.

Edit. Here is, what I think, the right question in the context of holomorphic differentials and hyperbolic metrics. For a fixed compact hyperbolic Riemann surface $X$ of genus $p$ let $L(X)$ denote the smallest bi-Lipschitz constant of identify maps $$ (X,h)\to (X,g), $$ where $h$ is the hyperbolic metric on $X$, the infimum is taken over all Riemannian metrics $g=\sum_{k=1}^p \omega_k \bar\omega_k$, where $\{\omega_1,...,\omega_p\}$ are bases in $\Omega^1(X)$. (Note that the identity map is conformal with respect to the metrics $h$ and $g$.) Let $\mathcal M_p$ denote the moduli space of compact Riemann surfaces of genus $p$.

Question. (1) Is $$ L_p:=\sup_{X\in {\mathcal M}_p} L(X) $$ finite? (2) If it is finite, what are the asymptotics of $L_p$ as $p\to\infty$?

I do not know how to answer this question and I do not remember seeing any work on this. (Clearly, for getting finiteness it suffices to analyze the behavior of degenerating families of genus $p$ Riemann surfaces. One should probably also analyze the case of Riemann surfaces of finite type so that one can run induction arguments. My guess is that $L_p$ is always infinite.)

First of all, the constants $c_i$ will have to depend on the complex structure of $X$ since without prescribing a complex structure one cannot talks about dependence on a basis of the space of holomorphic differentials on $X$. Now, $h$ and $g$ are both Riemannian metrics on a compact manifold and, thus, the identity map $$ (X,g)\to (X,h) $$ is bi-Lipschitz with Lipschitz constants depending, of course, on $g$ and $h$. Suppose that $(X, h), (Y, h)$ are Riemannian manifolds and $f: (X, g)\to (Y, h)$ is an $L$-Lipschitz map. Then for every closed geodesic $c$ in $(X,g)$, the $h$-length $\ell_h(c)$ of $f(c)$ is at most $L\ell_g(c)$. Hence, if $c^*$ is a shortest closed geodesic in the free homotopy class of $f(c)$, we have $$ \ell_h(c^*)\le L\ell_g(c). $$ For $L$-bi-Lipschitz maps one obtains the reverse inequality by considering $f^{-1}$. Thus, for every free homotopy class $[c]$ in $X$ you obtain the inequality $$ L^{-1}\ell_g([c]) \le \ell_h([f(c)])\le L \ell_g([c]), $$ where $\ell_\bullet([\bullet])$ stands for the length of a shortest loop in the given free homotopy class.

First of all, the constants $c_i$ will have to depend on the complex structure of $X$ since without prescribing a complex structure one cannot talks about dependence on a basis of the space of holomorphic differentials on $X$. Now, $h$ and $g$ are both Riemannian metrics on a compact manifold and, thus, the identity map $$ (X,g)\to (X,h) $$ is bi-Lipschitz with Lipschitz constants depending, of course, on $g$ and $h$. Suppose that $(X, g), (Y, h)$ are Riemannian manifolds and $f: (X, g)\to (Y, h)$ is an $L$-Lipschitz map. Then for every closed geodesic $c$ in $(X,g)$, the $h$-length $\ell_h(c)$ of $f(c)$ is at most $L\ell_g(c)$. Hence, if $c^*$ is a shortest closed geodesic in the free homotopy class of $f(c)$, we have $$ \ell_h(c^*)\le L\ell_g(c). $$ For $L$-bi-Lipschitz maps one obtains the reverse inequality by considering $f^{-1}$. Thus, for every free homotopy class $[c]$ in $X$ you obtain the inequality $$ L^{-1}\ell_g([c]) \le \ell_h([f(c)])\le L \ell_g([c]), $$ where $\ell_\bullet([\bullet])$ stands for the length of a shortest loop in the given free homotopy class.

Edit. Here is, what I think, the right question in the context of holomorphic differentials and hyperbolic metrics. For a fixed compact hyperbolic Riemann surface $X$ of genus $p$ let $L(X)$ denote the smallest bi-Lipschitz constant of identify maps $$ (X,h)\to (X,g), $$ where $h$ is the hyperbolic metric on $X$, the infimum is taken over all Riemannian metrics $g=\sum_{k=1}^p \omega_k \bar\omega_k$, where $\{\omega_1,...,\omega_p\}$ are bases in $\Omega^1(X)$. (Note that the identity map is conformal with respect to the metrics $h$ and $g$.) Let $\mathcal M_p$ denote the moduli space of compact Riemann surfaces of genus $p$.

Question. (1) Is $$ L_p:=\sup_{X\in {\mathcal M}_p} L(X) $$ finite? (2) If it is finite, what are the asymptotics of $L_p$ as $p\to\infty$?

I do not know how to answer this question and I do not remember seeing any work on this. (Clearly, for getting finiteness it suffices to analyze the behavior of degenerating families of genus $p$ Riemann surfaces. One should probably also analyze the case of Riemann surfaces of finite type so that one can run induction arguments. My guess is that $L_p$ is always infinite.)

Source Link
Moishe Kohan
  • 12.3k
  • 1
  • 36
  • 59

First of all, the constants $c_i$ will have to depend on the complex structure of $X$ since without prescribing a complex structure one cannot talks about dependence on a basis of the space of holomorphic differentials on $X$. Now, $h$ and $g$ are both Riemannian metrics on a compact manifold and, thus, the identity map $$ (X,g)\to (X,h) $$ is bi-Lipschitz with Lipschitz constants depending, of course, on $g$ and $h$. Suppose that $(X, h), (Y, h)$ are Riemannian manifolds and $f: (X, g)\to (Y, h)$ is an $L$-Lipschitz map. Then for every closed geodesic $c$ in $(X,g)$, the $h$-length $\ell_h(c)$ of $f(c)$ is at most $L\ell_g(c)$. Hence, if $c^*$ is a shortest closed geodesic in the free homotopy class of $f(c)$, we have $$ \ell_h(c^*)\le L\ell_g(c). $$ For $L$-bi-Lipschitz maps one obtains the reverse inequality by considering $f^{-1}$. Thus, for every free homotopy class $[c]$ in $X$ you obtain the inequality $$ L^{-1}\ell_g([c]) \le \ell_h([f(c)])\le L \ell_g([c]), $$ where $\ell_\bullet([\bullet])$ stands for the length of a shortest loop in the given free homotopy class.